Question

HELP ME ASAP PLEASE ITS THE SAME CONCEPT

Answer 1

• m∠Q = 133°

⠀

Explanation :

We know that,

• Sum of all angles of a triangle = 180°.

⠀

Therefore,

⠀

`{ \longrightarrow \sf \qquad \: m \angle \: P +m \angle Q + m\angle R = 180 {}^( \circ) }`

`{ \longrightarrow \sf \qquad \: {(x + 13)}^( \circ) +{(10x + 13)}^( \circ) + {(2x - 2)}^( \circ) = 180 {}^( \circ) }`

⠀

Adding like terms we get :

`{ \longrightarrow \sf \qquad \: {(x + 10x + 2x)} +(13^( \circ) + 13^( \circ) - 2^( \circ) ) = 180 {}^( \circ) }`

`{ \longrightarrow \sf \qquad \: 13x +24^( \circ) = 180 {}^( \circ) }`

`{ \longrightarrow \sf \qquad \: 13x = 180 {}^( \circ) - 24^( \circ) }`

`{ \longrightarrow \sf \qquad \: 13x = 156 {}^( \circ) }`

`{ \longrightarrow \sf \qquad \: x = \frac{156 {}^( \circ) }{13} }`

`{ \longrightarrow \qquad \: { \pmb{ x = 12 {}^( \circ) } }}`

⠀

Therefore,

• The value of x = 12°

⠀

Now, Substituting the value of x in m∠Q :

⠀

`{ \longrightarrow \qquad \: { \sf{ m \angle Q =( 10x + 13) {}^( \circ) } }}`

`{ \longrightarrow \qquad \: { \sf{ m \angle Q =10(12) {}^( \circ) + 13{}^( \circ) } }}`

`{ \longrightarrow \qquad \: { \sf{ m \angle Q =120 {}^( \circ) + 13{}^( \circ) } }}`

`{ \longrightarrow \qquad \: { \pmb{ m \angle Q =133 {}^( \circ) } }}`

⠀